Calculating Crate Acceleration: Force, Friction, and Net Force

[Larrytsai]

A 210-kg crate is pushed horizontally with a force of 730N. If the coefficient of the kinetic friction is 0.19, calculate the acceleration of the crate.

Attempt: FN= 2058

Ff=0.19(391)
Force of friction= 391N

Fnet= Fpush-Ff

ma= Fpush - Ff
2058a = 730 - 391

a=339/2058

can anyone check if my answers right, and if I am wrong explain how to do it?

 

[cryptoguy]

ok, first, i assume this is just a typo but you wrote Ff=0.19(391), but Ff = .19*2058, which i assume is 391.
now more importantly, you say "ma= Fpush - Ff" which is correct but "2058a = 730 - 391" is not. Remember it's m*a for the left part, while you have FN*a

 

[physixguru]

Simple method:

Force by push - force by friction = mass of block *acceleration.

730-.19*(210*9.8) = 210 * a
>338.98= 210 * a
>a= 338.98/210= 1.6141 m/s^2